Union of $l^p$, 0Is it true that $$ \bigcup\limits_{0<p<1} l^p = l^1\quad?$$ 
The space $l^p$ is the space of the sequences $\{a_n\}_n$ with $\sum |a_n|^p <\infty.$
The one inclusion is obvious, as any function in $l^p$ with $p<1$ also belongs in $l^1$. So the question is whether there exist a sequence in $l^1$ such that it doesn't belong in any $l^p$ with p<1. 
 A: Try $(a_n)_n$ defined by
$$
a_{n} = \frac{1}{n \ln^2 n}
$$
for $n\geq2$ (and arbitrary for $a_0,a_1$).

Note: the series of the form $\sum_n \frac{1}{n^\alpha \ln^\beta n}$ are called Bertrand series ("séries de Bertrand" in French) and converge if, and only if, $\alpha > 1$ or ($\alpha=1$ and $\beta > 1$).
A: I have a non-constructive proof.
First, I claim that this equality in sets implies the following equality
$$\bigcup_{0 < p < 2} \ell^p = \ell^2.$$
Note that, as with the previous equality, the $\subseteq$ inclusion is trivial. Suppose $\bigcup_{0 < p < 1} \ell^p = \ell^1$ and $(x_n) \in \ell^2$. Then $\sqrt{|x_n|} \in \ell^1$, and hence $\sqrt{|x_n|} \in \ell^p$ for some $p < 1$. Thus, $|x_n|$ and hence $x_n$ are in $\ell^{2p}$, where $2p < 2$, as required.
Now, we can further refine this union to the countable union,
$$\bigcup_{n \in \Bbb{N}} \ell^{2 - 1/n} = \ell^2.$$
Note that $\ell^{2 - 1/n} \subsetneq \ell^{2 - 1/(n+1)}$ for all $n$, and form a nested sequence of Banach subspaces of $\ell^2$. All of these subspaces are complete, and hence are closed in $\ell^2$.
By Baire Category theorem, at least one of the subspaces must have non-empty interior. The only subspace to have non-empty interior is the full space, in other words, there exists some $n$ such that
$$\ell^{2 - 1/n} = \ell^2,$$
which is plainly false by examining $p$-series. Thus, the set equalities both do not hold.
